A particle of mass m is executing oscillations about the origin on the $$x-axis$$. Its potential energy is $$U(x)$$ $$=$$ k[x]$$3$$, where k is a positive constant. If the amplitude of oscillation is a, then its time period T is

Question

A particle of mass m is executing oscillations about the origin on the $$x-axis$$. Its potential energy is $$U(x)$$ $$=$$ k[x]$$3$$, where k is a positive constant. If the amplitude of oscillation is a, then its time period T is

Infinity Learn · Accepted Answer

U $$=$$ k|x|$$3$$  ⇒ F $$=$$ $$-dUdx$$ $$=$$ $$-3k$$|x|$$2----------------(i)Also$$, for SHM x $$=$$ a $$sin$$ ωt and $$d2xdt2+$$ω$$2x=$$ $$0$$⇒acceleration $$=$$ $$d2xdt2$$ $$=$$ $$-$$ω$$2x$$  ⇒ F $$=$$ $$ma=$$ $$md2xdt2$$ $$=$$ $$-m$$ω$$2x----(ii)From$$ equation (i) and (ii) we get ω $$=$$ $$3kxm$$⇒T $$=$$ $$2$$πω $$=$$ $$2$$$$π$$$$m3kx$$ $$=$$ $$2$$$$π$$$$m3k(a$$ $$sin$$ ω$$t)$$ ⇒ T ∝ $$1a$$

A particle of mass m is executing oscillations about the origin on the x-axis. Its potential energy is $U (x) = k {[x]}^{3},$ where k is a positive constant. If the amplitude of oscillation is a, then its time period T is

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A particle of mass m is executing oscillations about the origin on the x-axis. Its potential energy is U(x) = k[x]3, where k is a positive constant. If the amplitude of oscillation is a, then its time period T is

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A particle of mass m is executing oscillations about the origin on the x-axis. Its potential energy is $U (x) = k {[x]}^{3},$ where k is a positive constant. If the amplitude of oscillation is a, then its time period T is